Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A friend says that the faster the reaction, the larger the equilibrium constant. Is your friend correct? Why or why not? [Sections \(15 \cdot 1\) and 15.2\(]\)

Short Answer

Expert verified
Your friend is incorrect. The rate of reaction and the equilibrium constant are separate aspects of a chemical reaction and do not have a direct relationship. The rate of reaction measures how fast the system reaches equilibrium, while the equilibrium constant describes the balanced state of the system. A faster reaction does not necessarily correspond to a larger equilibrium constant, as the equilibrium constant depends on the specific nature and properties of the reactants and products involved in the reaction.

Step by step solution

01

Understanding rate of reaction

Rate of reaction refers to the speed at which a chemical reaction occurs. It is determined by several factors, including concentration, temperature, presence of a catalyst, and the nature of the reactants.
02

Understanding equilibrium constant

The equilibrium constant (K) is a numerical value that describes the ratio of concentrations of products to reactants in a chemical reaction when the system reaches a state of equilibrium. Equilibrium is the state in which both reactants and products are present at constant concentrations.
03

Difference between rate of reaction and equilibrium constant

While both the rate of reaction and equilibrium constant describe aspects of a chemical reaction, they are not directly related. The rate of reaction is a measure of how quickly reactants are turned into products, whereas the equilibrium constant tells us the relative concentrations of reactants and products when the system reaches equilibrium. In other words, the rate of reaction describes how fast the system reaches equilibrium, while the equilibrium constant describes the balanced state of the system.
04

Relationship between rate of reaction and equilibrium constant

Since the rate of reaction measures how fast the system reaches equilibrium and the equilibrium constant describes the balanced state, it is not appropriate to say that a faster reaction always corresponds to a larger equilibrium constant. The equilibrium constant depends on the specific nature and properties of the reactants and products involved in the reaction, not just its rate. Therefore, the statement made by the friend is incorrect. To conclude, your friend's statement that the faster the reaction, the larger the equilibrium constant is incorrect. The rate of reaction and the equilibrium constant are separate aspects of a chemical reaction and do not have a direct relationship as described by your friend.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rate of Reaction
The rate of reaction tells us how fast or slow a chemical reaction occurs. It is the speed at which reactants transform into products in a given amount of time.
Several factors influence the rate of reaction:
  • Concentration: More molecules in a space means a higher chance of collisions, which can speed up a reaction.
  • Temperature: Higher temperatures increase kinetic energy, causing molecules to collide more frequently and forcefully.
  • Presence of a Catalyst: Catalysts provide an alternative pathway with lower activation energy for the reaction, speeding it up.
  • Nature of Reactants: Substances react at different speeds based on their physical and chemical properties.
These factors determine how quickly a reaction approaches equilibrium, but they don't tell us anything about the equilibrium state itself.
Equilibrium Constant
The equilibrium constant (denoted as \( K \)) expresses the ratio of concentrations of products to reactants at equilibrium for a reversible chemical reaction.
The equilibrium constant is defined by the equation:\[K = \frac{{[Products]}}{{[Reactants]}}\]Where \([Products]\) and \([Reactants]\) are the equilibrium concentrations of products and reactants, respectively.
This value remains constant at a given temperature, reflecting the relative stability of the reactants and products.
The equilibrium state means that the rate of the forward reaction equals the rate of the reverse reaction, making the concentrations of reactants and products unchanging over time.
Importantly, \( K \) is independent of the reaction rate and solely depends on the environmental conditions and the properties of the reactants and products themselves.
This explains why it is incorrect to associate a large equilibrium constant with a fast reaction.
Chemical Kinetics
Chemical kinetics is the branch of chemistry that studies the rates of chemical processes and the factors affecting them.
It provides insight into reaction mechanisms by exploring how different variables influence reaction speed.
Key considerations in chemical kinetics include:
  • Reaction Rate: Examining how quickly reactants convert into products.
  • Mechanism Pathways: Understanding detailed steps that reactions follow from start to finish.
  • Rate Laws: Mathematical expressions that represent the relationship between reaction rate and concentrations of reactants.
Unlike equilibrium studies, which focus on the state where reactions stop changing, chemical kinetics deals with the dynamic processes leading up to that state.
This field allows chemists to manipulate conditions to control reaction speeds, providing invaluable information for industrial processes, laboratory practices, and more.
Through chemical kinetics, we can see why your friend’s assertion about reaction speed and equilibrium constants doesn’t hold true; while they study interconnected aspects of reactions, they focus on fundamentally different properties.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Explain why we normally exclude pure solids and liquids from equilibrium- constant expressions.

\(\mathrm{NiO}\) is to be reduced to nickel metal in an industrial process by use of the reaction $$\mathrm{NiO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(s)+\mathrm{CO}_{2}(g)$$ At \(1600 \mathrm{~K}\) the equilibrium constant for the reaction is \(K_{p}=6.0 \times 10^{2}\). If a CO pressure of 150 torr is to be employed in the furnace and total pressure never exceeds 760 torr, will reduction occur?

At \(1000 \mathrm{~K}, K_{p}=1.85\) for the reaction $$ \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g) $$ (a) What is the value of \(K_{p}\) for the reaction \(\mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) ?(\mathbf{b})\) What is the value of \(K_{p}\) for the reaction \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\) ? (c) What is the value of \(K_{c}\) for the reaction in part (b)?

A mixture of \(\mathrm{H}_{2}, \mathrm{~S},\) and \(\mathrm{H}_{2} \mathrm{~S}\) is held in a \(1.0-\mathrm{L}\) vessel at \(90{ }^{\circ} \mathrm{C}\) and reacts according to the equation: $$\mathrm{H}_{2}(g)+\mathrm{S}(s) \rightleftharpoons \mathrm{H}_{2} \mathrm{~S}(g)$$ At equilibrium the mixture contains \(0.46 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{~S}\) and \(0.40 \mathrm{~g}\) \(\mathrm{H}_{2}\) (a) Write the equilibrium-constant expression for this reaction. (b) What is the value of \(K_{c}\) for the reaction at this temperature? (c) Why can we ignore the amount of \(\mathrm{S}\) when doing the calculation in part (b)?

Consider the reaction \(\mathrm{IO}_{4}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) \(\mathrm{H}_{4} \mathrm{IO}_{6}^{-}(a q) ; K_{c}=3.5 \times 10^{-2} .\) If you start with \(25.0 \mathrm{~mL}\) of a \(0.905 \mathrm{M}\) solution of \(\mathrm{NaIO}_{4}\), and then dilute it with water to \(500.0 \mathrm{~mL}\), what is the concentration of \(\mathrm{H}_{4} \mathrm{IO}_{6}^{-}\) at equilibrium?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free